Racetovb4j

Answered

2022-02-03

The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=-3, how do you find a possible formula for P(x)?

Answer & Explanation

bubble53zjr

Expert

2022-02-04Added 14 answers

Since each root is a linear factor, we can write:

$P\left(x\right)={x}^{2}{(x-1)}^{2}(x+3)$

$={x}^{2}({x}^{2}-2x+1)(x+3)$

$={x}^{5}+{x}^{4}-5{x}^{3}+3{x}^{2}$

Any polynomial that contains these zeros and at least these multiplicities is going to be a multiple (scalar or polynomial) of this P(x)

Answer:

$P\left(x\right)={x}^{5}+{x}^{4}-5{x}^{3}+3{x}^{2}$

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